Mathematical Conundrum

[Shnoz (Sophie)]

I have chanced upon a mathematical paradox in an equation known as the Lorentz transformation, and I thought this would probably be a good place to find an answer.
Basically from my understanding the Lorentz transformation gives an observer's time compared to a moving object's time, because moving object's have slower times. I've tested the equation with all kinds of velocities and it always gives back a logical result, except for when I put in the velocity of light!
The equation is as follows.
't = (t - v/csquared)/(square root of1 - vsquared/csquared)
't - observer's time
t - moving object's time
v - velocity
c - speed of light
Hopefully it's possible to understand my write out of the equation.
So if the velocity were to be the speed of light then we would eventually arrive at a division by 0, which is supposed to be impossible. So what exactly is the answer to this equation? What's the time difference between and observer's time and a photon's time?
Or I could have simply made a mistake in my understanding. Either way, I need answers!

[avandonk]

If a photon had consciousness it would perceive itself travelling across the Universe in an instant! Scary stuff.

Bert

[NQLD_Newby]

I could be corrected, but my understanding is, that light travels at "C" because it is massless. Furthermore anything with mass cannot travel at or above the speed of light. This is why your formula doesn't work for the speed of light.

If you use V=C in your calculation the C simply cancels itself out. Another thing, the square root of 1 is 1.

t=(t-V/Csquared)/(square root of 1-Vsquared/Csquared)

=(t-C/C*C)/(Square root of 1-C*C/C*C)

=(t-C)/(Square root of1-1)

=(t-c)/(1-1)

=(t-C)/0

=0

So I guess this formula only works for objects which have mass.

[sjastro]

Originally Posted by NQLD_Newby

=(t-c)/(1-1)

=(t-C)/0

=0

(t-C)/0 is undefined.

Regards

Steven

[sjastro]

Originally Posted by Shnoz

I have chanced upon a mathematical paradox in an equation known as the Lorentz transformation, and I thought this would probably be a good place to find an answer.
Basically from my understanding the Lorentz transformation gives an observer's time compared to a moving object's time, because moving object's have slower times. I've tested the equation with all kinds of velocities and it always gives back a logical result, except for when I put in the velocity of light!
The equation is as follows.
't = (t - v/csquared)/(square root of1 - vsquared/csquared)
't - observer's time
t - moving object's time
v - velocity
c - speed of light
Hopefully it's possible to understand my write out of the equation.
So if the velocity were to be the speed of light then we would eventually arrive at a division by 0, which is supposed to be impossible. So what exactly is the answer to this equation? What's the time difference between and observer's time and a photon's time?
Or I could have simply made a mistake in my understanding. Either way, I need answers!

The Lorentz transformations are only applicable for inertial frames of references travelling at speeds of less than C.

It follows the basic postulate of special relativity that no mass particle can attain the speed of light.

It's debatable if one can even assign a frame of reference to a photon. A photon can be simultaneously at the start and end of it's journey.

Regards

Steven

[sheeny (Al)]

Originally Posted by Shnoz

The equation is as follows.
't = (t - v/csquared)/(square root of1 - vsquared/csquared)
't - observer's time
t - moving object's time
v - velocity
c - speed of light
So what exactly is the answer to this equation? What's the time difference between and observer's time and a photon's time?
Or I could have simply made a mistake in my understanding. Either way, I need answers!

I'm not sure about this equation Sophie. It doesn't look like the time dilation formula I'm used to. See:

http://en.wikipedia.org/wiki/Time_dilation

But no matter... The faster something travels, the more it slows down. The limit of the formula as v approaches c is 0 (even though technically it is undefined, the limit provides some guidance on what to expect). So as Bert already replied, the photon doesn't experience time... it "experiences" going from start to finish instantly.



Al.

[sjastro]

Originally Posted by sheeny

I'm not sure about this equation Sophie. It doesn't look like the time dilation formula I'm used to. See:

http://en.wikipedia.org/wiki/Time_dilation

But no matter... The faster something travels, the more it slows down. The limit of the formula as v approaches c is 0 (even though technically it is undefined, the limit provides some guidance on what to expect). So as Bert already replied, the photon doesn't experience time... it "experiences" going from start to finish instantly.



Al.

The time dilation formula is not a Lorentz transformation even though it contains the Lorentz contraction factor.

The formula quoted by Sophie has an x missing. (x is the distance travelled along the x axis).

It should read

't = (t - (v/csquared)x)/(square root of1 - vsquared/csquared)

As v approaches c, 't approaches infinity.

Regards

Steven

[sculptor]

I'm an imager, not a relativist, so add salt.

The concept of the photon's departure event and arrival event having the same time coordinate in the photon's reference frame seems very useful.

In the two-slit experiment, for the photon, there is no WHILE. It is at the source, the open slit, the closed slit, and the screen, all at the same time.

I would dearly like to see a decent explanation of the arcane variants of the two-slit experiment, written "in this light", if you'll excuse the pun.

Explanations "in this light" may also help with dismissing possibly content-free statements like "SN1987a occurred millions of years ago".

[xelasnave]

Steven said....
"A photon can be simultaneously at the start and end of it's journey"

Steven can I trouble you for more help.

Does this mean a photon that left "the other end of the Universe" to arrive here is at both ends of its journey simultaneously?
or does it mean that the photo only "thinks" it is?

Or what I should say this seems strange is it possible to explain this to a lay man so it makes sence.

Anyways I have followed this thread with absolute interest and am trying to get my head around it all.


Alex

[xelasnave]

Originally Posted by avandonk

If a photon had consciousness it would perceive itself travelling across the Universe in an instant! Scary stuff.

Bert

Bert I guess this is along the lines of the question I asked Steven...
Is this for real or stuff that the math suggests or simply something that because I dont understand math very well that will always be beyond me.

alex

[sjastro]

Originally Posted by xelasnave

Steven said....
"A photon can be simultaneously at the start and end of it's journey"

Steven can I trouble you for more help.

Does this mean a photon that left "the other end of the Universe" to arrive here is at both ends of its journey simultaneously?
or does it mean that the photo only "thinks" it is?

Or what I should say this seems strange is it possible to explain this to a lay man so it makes sence.

Anyways I have followed this thread with absolute interest and am trying to get my head around it all.


Alex

Hello Alex.

The best way to explain this is with a thought experiment.

Suppose I'm driving in my car and I turn my headlights on. You're at a location a few kilometers down the road. The photons from my headlights reach you after a time t= d/c where d is the distance from you when I turned my headlights on and c is the speed of light.

Suppose there are other observers at various locations along the road. They receive the photons after t1=d1/c, t2= d2/c, t3= d3/c etc.

Note that all observers will measure the speed of the photon as c which is independant of the speed of my car as I approach the observers.

Since my car is travelling very much slower than c each observer will note that I turn my headlights on followed by a period of elapsed time when I drive past their locations. The elapsed time is t(1)= d1/u, t(2)=d2/u, t(3)=d3/u etc where u is the speed of my car.

This order of events is consistant with each observer.

Now let's suppose I defy SR and my car is now travelling at u=c. I turn my headlights on. What happens now. Both the car and the photons arrive at the same time for each observer. Each observer will claim I turned my headlights on at the instant I passed their location despite the fact that each observer is in a different location.

At the speed of light all events become simultaneous.

There is a third scenario. What happens if a photon is able to pass information faster than the speed of light? It can be shown mathematically that causality is violated. Some observers will see event A happen before event B, others will see event B occur before event A.

Hope this is of help.

Regards

Steven

[xelasnave]

Thank you Steven I really appreciate you help

alex

[Shnoz (Sophie)]

Thanks everyone for the help with that equation. I've learnt quite a few things. I didn't realise that photons must experience an 'instantaneous' travel between point A and B, it is an interesting concept.

[NQLD_Newby]

Originally Posted by sjastro

Hello Alex.

The best way to explain this is with a thought experiment.

Suppose I'm driving in my car and I turn my headlights on. You're at a location a few kilometers down the road. The photons from my headlights reach you after a time t= d/c where d is the distance from you when I turned my headlights on and c is the speed of light.

Suppose there are other observers at various locations along the road. They receive the photons after t1=d1/c, t2= d2/c, t3= d3/c etc.

Note that all observers will measure the speed of the photon as c which is independant of the speed of my car as I approach the observers.

Since my car is travelling very much slower than c each observer will note that I turn my headlights on followed by a period of elapsed time when I drive past their locations. The elapsed time is t(1)= d1/u, t(2)=d2/u, t(3)=d3/u etc where u is the speed of my car.

This order of events is consistant with each observer.

Now let's suppose I defy SR and my car is now travelling at u=c. I turn my headlights on. What happens now. Both the car and the photons arrive at the same time for each observer. Each observer will claim I turned my headlights on at the instant I passed their location despite the fact that each observer is in a different location.

At the speed of light all events become simultaneous.

Hi everyone, I love this stuff.

Steven could I ask for some further clarification please?

I understand your statement, however, if t1, t2 and t3 above are measureable in your first example with the slow moving car, wouldn't the three observers simply claim that you turned your headlights on at a different time? (observer 1 @t1, ob2 @t2, and so on). I realise that you actually turned them on once only, but isn't that what relativity is all about? Everyone will have different measurements of such experiments because their measurements are relative to their position in time-space? This includes your own perceptions of when you turned the lights on.

Please bare with me, and correct me if I'm wrong as I am only just learning all this stuff. My statements are mearly how I understand GR, and SR, and I would love some further guidance where this is concerned.

[sjastro]

Originally Posted by NQLD_Newby

Hi everyone, I love this stuff.

Steven could I ask for some further clarification please?

I understand your statement, however, if t1, t2 and t3 above are measureable in your first example with the slow moving car, wouldn't the three observers simply claim that you turned your headlights on at a different time? (observer 1 @t1, ob2 @t2, and so on). I realise that you actually turned them on once only, but isn't that what relativity is all about? Everyone will have different measurements of such experiments because their measurements are relative to their position in time-space? This includes your own perceptions of when you turned the lights on.

Please bare with me, and correct me if I'm wrong as I am only just learning all this stuff. My statements are mearly how I understand GR, and SR, and I would love some further guidance where this is concerned.

Hello Rex.

Observers 1,2,3... are all stationary and are therefore in the same frame of reference.

t1,t2, t3...... are elapsed times and varies according to distance. Since each observer is in the same frame, the time when the headlights are turned on is synchronized for all observers.

If the observers are in motion at different velocites, then the time is no longer synchronized. The time for each observer is calculated by the Lorentz transformation.

Regards

Steven

[NQLD_Newby]

Hmmmm.... I must be misunderstanding something then.

Let me try to explain what I mean, in a hope that you can help me see where I am going wrong.


There are three observers, 1, 2, and 3.

Obs1 is 10 light minutes away from you, obs 2 is 20 light minutes away and obs 3 is 30 light minutes away.

You are travelling at the speed of light, and at exactly 10.00am you turn on your lights.

Because you are travelling at the speed of light, none of the observers can see you until you pass them. They cannot see your lights until then either as you and the photons are travelling at the same speed.

It takes 10 minutes for you and the photons from your lights to reach obs1, therefore he says you turned your lights on at 10.10am.

It takes 20 minutes to reach obs2 so he says 10.20am is the time you turned them on.

Likewise obs3 says 10.30am is when you turned them on.

So between you and the three observers you have four different times for the same event. Each of you is correct, as your measurements were made relative to your position in time-space.

The difference is, similar to the way the deeper into space you look, the further back in time you go, obs1 is looking 10 minutes back in time obs2 20 minutes and obs3 is looking 30 minutes back in time. But relative to their positions the event happenned at the times they have stated.

This is how I understood the process, but would like to get it right, so please point me in the required direction.

[Sionnagh (Mick)]

Don't forget that when you turn your lights on if you were to measure the speed of the light then you would see it moving away from you at the speed of light...


Mick

[NQLD_Newby]

Hi Mick,

Once again please feel free to correct me if I am missing something, but the speed of light is always the speed of light, regardless of your speed. Meaning the speed of light doesn't add itself to your own speed. If you were moving at 100km/hr and turn your lights on, the light coming from your car isn't C+100km/hr it is still C. Therefore if you are moving at the speed of light, you wouldn't see any light at all because it cannot get out in front of you because you are moving at the same speed.

In fact to be perfectly correct, in the scenario described in the above posts, the observers wouldn't see your lights come on at all, because you and the light from your headlights would arrive at exactly the same time.......I think?

[sjastro]

Originally Posted by NQLD_Newby

Hmmmm.... I must be misunderstanding something then.

Let me try to explain what I mean, in a hope that you can help me see where I am going wrong.


There are three observers, 1, 2, and 3.

Obs1 is 10 light minutes away from you, obs 2 is 20 light minutes away and obs 3 is 30 light minutes away.

You are travelling at the speed of light, and at exactly 10.00am you turn on your lights.

Because you are travelling at the speed of light, none of the observers can see you until you pass them. They cannot see your lights until then either as you and the photons are travelling at the same speed.

It takes 10 minutes for you and the photons from your lights to reach obs1, therefore he says you turned your lights on at 10.10am.

It takes 20 minutes to reach obs2 so he says 10.20am is the time you turned them on.

Likewise obs3 says 10.30am is when you turned them on.

So between you and the three observers you have four different times for the same event. Each of you is correct, as your measurements were made relative to your position in time-space.

Rex,

What you're describing is the elapsed time, not the time ordinate as defined by a specific event in space time. An event in space time is defined by the coordinates (x,y,z,t).

Since the three observers are in the same frame of reference, they agree on a synchronized time as to when the headlights were turned on. In other words the t value is the same for each observer.

Plugging in v=0 (since each observer is stationary) in
t' = (t - (v/csquared)x)/(square root of1 - vsquared/csquared)

Gives t'=t which proves the event is synchronized.

Suppose this time is 10.00am. If the car travels at the speed of light, the headlights are turned on at each location of the observer (as we both agree to). Each observer however has noted that the lights are turned on at 10.00am and therefore concludes the car has passed their locations at 10.00am.

Compare this when the car is travelling at more conventional speeds.
The time when the car passes each location is simply 10.00am plus the time it takes for the car to reach each destination.

The difference is, similar to the way the deeper into space you look, the further back in time you go, obs1 is looking 10 minutes back in time obs2 20 minutes and obs3 is looking 30 minutes back in time. But relative to their positions the event happened at the times they have stated.

What it means is that the signal for the event has taken 10, 20 or 30 minutes to reach the observers. If all the observers are travelling at the same velocity (<< c) or are stationary you get the same synchronized result irrespective of the distance of the observers from the source.

If the observers are travelling at different velocites (and at high velocities for the effects to be noticeable), the event is no longer synchronized.

Hope this is useful.

Regards

Steven

[NQLD_Newby]

Hang on I think I've got it..............?????

Using my own explanation above........... When you reach obs1 all the photons from 10min worth of having your lights on would reach him at 10.10am. therefore he would see you at the point where you turned your lights on at 10.10am.....but he would also see you at each and every point along your path of travel at 10.10am. Therefore you would appear to be in each and every position along your path all at once, in other words you would be an instantanious blur from the point at which you turned your lights on until you passed the observer. The other two observers would see similar effects but obs2 would see 20min of your path, and obs3 would see 30min. The only difference would be the time at which they said the event happenned, ie. 10min apart.

Man this stuff really gets your brain working.

I'm not really sure about what I have said here but I think you can get the idea of where I'm coming from.