If I could travel at the speed of light

[Paduan (Brett)]

the faster that you travel, time apparently travels slower from .9 C onwards anyway. So at the speed of light you could be every where yet nowhere. all of your friends not with you will age roughly 100 yrs per your day. not to mention that at 10g (9.8 m/s/s) it would take roughly 5 yrs to get to such a velocity anyway... but remember that the speed of sound was impossible not 50 yrs ago


P.S. do not quote me on the numbers as i am rounding from memory for "shock" value

[leonie-macaroni (Leonie)]

Originally Posted by Paduan

all of your friends not with you will age roughly 100 yrs per your day.

But you would be massless anyway, and nobody likes massless friends.
Photons and gluons pass me by constantly and I mostly just ignore them.

I suppose they could be friends with each other, but as stated earlier that theoretically travelling at C will cause no time for the traveller to pass at all, I would guess that they haven't met yet.

[Mliss (Mel)]

Thanks for posting that Glen.

For a noob that's about as much as i can handle
Puts a real perspective on how tiny we are, and how much space is in.. space!

I can't keep up with you guys and my head can't cope with time dilation, redshift and spacetime curvature :exploding head emoticon:

[renormalised (Carl)]

Originally Posted by Paduan

the faster that you travel, time apparently travels slower from .9 C onwards anyway. So at the speed of light you could be every where yet nowhere. all of your friends not with you will age roughly 100 yrs per your day. not to mention that at 10g (9.8 m/s/s) it would take roughly 5 yrs to get to such a velocity anyway... but remember that the speed of sound was impossible not 50 yrs ago


P.S. do not quote me on the numbers as i am rounding from memory for "shock" value

Actually, the moment you begin to move, time starts to slow down for you, relative to an observer at rest to you. It's imperceptibly changed at first, but the closer you come to "c", the slower it becomes.

At the speed of light your own time, from your perspective, remains the same, but a stationary clock's time speeds up to the point where, theoretically, all of time outside of your ship passes almost instantaneously. It's the same effect you get when you reach and pass over the vent horizon of a black hole. The opposite effect happens for the stationary observer looking at your clock. At "c", you clock from their perspective, stops. Same when an object approaches and passes over the EH of a BH. From your PoV, it never crosses. That's time dilation for you

[renormalised (Carl)]

Originally Posted by leonie-macaroni

But you would be massless anyway, and nobody likes massless friends.
Photons and gluons pass me by constantly and I mostly just ignore them.

I suppose they could be friends with each other, but as stated earlier that theoretically travelling at C will cause no time for the traveller to pass at all, I would guess that they haven't met yet.

Actually, if you were a photon, gluon or graviton, you would be massless, but not for the reason you're thinking of. The velocity of the particle (in this case bosons, or force carrying particles) does not dictate the mass of the particle. That has to do with the way particles couple with the Higgs Field, and since these bosons don't couple to this field, they're massless...no matter (pardon the pun) what velocity they're doing. The only bosons that have any mass at all are the W and Z particles, which intermediate the weak nuclear force responsible for radioactive decay. They cannot travel at "c".

[glenc (Glen)]

Thanks Mel. This was intended to be for beginners.

[pmrid (Peter)]

But since light would also be travelling towards you at the same speed, would you seem to be travelling at twice that speed (relative to that light source)?
Peter

[renormalised (Carl)]

Originally Posted by pmrid

But since light would also be travelling towards you at the same speed, would you seem to be travelling at twice that speed (relative to that light source)?
Peter

No, the speed of light is a constant, no matter what situation you find yourself in. So if you were traveling at c and were being approached by a light source traveling at c in the opposite direction, your closing velocity would still be c, not 2c. Whilst traveling through space, the speed of light cannot be exceeded in any way, shape or form.

[pmrid (Peter)]

Originally Posted by renormalised

No, the speed of light is a constant, no matter what situation you find yourself in. So if you were traveling at c and were being approached by a light source traveling at c in the opposite direction, your closing velocity would still be c, not 2c. Whilst traveling through space, the speed of light cannot be exceeded in any way, shape or form.

The operative word there was "seem" or "relative" - or whatever I said. While light is not traveling faster than C, it appears to be - just like an approaching car on the highway. No??
And what about Doppler shift - I'm approaching so it should shift to blue. It's approaching me also, so it should also shift to blue. So in combination, the combined blue shift would put it outside the visual spectrum anyway - Yes/No/Maybe?
Peter

[renormalised (Carl)]

Originally Posted by pmrid

The operative word there was "seem" or "relative" - or whatever I said. While light is not traveling faster than C, it appears to be - just like an approaching car on the highway. No??

Light can only travel at c, whatever c is in the medium within which the light is traveling. Be it a vacuum or a bowl of jelly.

The answer to your question is still no....c is a constant. To change the equations to get 2c, you would have to change the constancy of the speed of light, which you can't do. As I mentioned before, your example doesn't produce a "2c" result, whether that's "seems" or "relative" to anything you or the other object is doing.

Originally Posted by pmrid

And what about Doppler shift - I'm approaching so it should shift to blue. It's approaching me also, so it should also shift to blue. So in combination, the combined blue shift would put it outside the visual spectrum anyway - Yes/No/Maybe?
Peter

You will appear to blueshift to it and it to you. And the answer to the second part of this question is No. You will only see it blueshifted by the amount it attains depending on its velocity of approach. There is no combination of "blueshifts" to give you an ultraviolet spectrum. You will only see the light of the object as it approaches you being blueshifted.

[Mliss (Mel)]

Glen, what have you started?

i'm really trying to grasp this, (don't like my chances) but it's a really interesting read! so many talented people in here!

[The_bluester (Paul)]

Originally Posted by adman

...and at that speed it wouldn't be pretty!

I am not so sure about that. Maybe it wouldn't be pretty, but it would at least be pretty spectacular!

[pmrid (Peter)]

Originally Posted by renormalised

Light can only travel at c, whatever c is in the medium within which the light is traveling. Be it a vacuum or a bowl of jelly.

The answer to your question is still no....c is a constant. To change the equations to get 2c, you would have to change the constancy of the speed of light, which you can't do. As I mentioned before, your example doesn't produce a "2c" result, whether that's "seems" or "relative" to anything you or the other object is doing.

OK. I'm struggling here.
Photon A and I both leave our respective destinations at the same time, heading towards one-another. At the point of departure we are, say, 5 light years apart. Our relative starting positions.

After 1 year, I have travelled a light year towards the source of photon A. In the same year, Photon A has travelled a light year in my direction.
At the end of year 1, we are therefore 3 light years apart, not 5. To my observing eye, hasn't the distance between us reduced at the rate of 2 light years/year?

Peter

[Brian W (Brian)]

And from another perspective when we look at those distant objects we are actually connected to an uninterrupted energy transmission that originated with the object we are looking at.

Which means that an energy source that includes me is also traveling at the speed of light.

As i have often said my telescope allows me to travel at the speed of light with no fear of a speeding ticket!
brian

[sjastro]

Originally Posted by pmrid

OK. I'm struggling here.
Photon A and I both leave our respective destinations at the same time, heading towards one-another. At the point of departure we are, say, 5 light years apart. Our relative starting positions.

After 1 year, I have travelled a light year towards the source of photon A. In the same year, Photon A has travelled a light year in my direction.
At the end of year 1, we are therefore 3 light years apart, not 5. To my observing eye, hasn't the distance between us reduced at the rate of 2 light years/year?

Peter

An understanding of special relativity is required here.

At very high velocities approaching and including the speed of light (=c), velocities don't add vectorially.

For example if two objects approach each other at a velocity v, then their approach velocity relative to each other is 2v.

At high velocities their approach velocity is 2v/(1+(v/c)^2)

If each object (photon) approaches each other at a speed c, then the
approach velocity is 2c/(1+(c/c)^2)= c

Regards

Steven

[pmrid (Peter)]

Originally Posted by sjastro

An understanding of special relativity is required here.

At very high velocities approaching and including the speed of light (=c), velocities don't add vectorially.

For example if two objects approach each other at a velocity v, then their approach velocity relative to each other is 2v.

At high velocities their approach velocity is 2v/(1+(v/c)^2)

If each object (photon) approaches each other at a speed c, then the
approach velocity is 2c/(1+(c/c)^2)= c

Regards

Steven

I am going to have to take that on faith alone. It sounds like fancy bookeeping to me I'm afraid.
Peter

[sjastro]

Originally Posted by pmrid

I am going to have to take that on faith alone. It sounds like fancy bookeeping to me I'm afraid.
Peter

The derivation of the formula is simple, straightforward and logical.

Regards

Steven

[scagman (John)]

Originally Posted by Mliss

Glen, what have you started?

i'm really trying to grasp this, (don't like my chances) but it's a really interesting read! so many talented people in here!

I've given up triing to grasp it, I have enough trouble keeping a grasp on reality.


So if I was on a really really big train(t1) travelling along the tracks at the speed of light, and inside that train there is another train(t2) travelling along a set of tracks running down the center of the really big train, also travelling at the speed of light, what speed is the inside train(t2) going in relation to someone watching from outside the big train?

Now to me it should be travelling at 2*c but I know its not because you cant go faster than c.

Now you could complicate matters further and have a really really small train(t3) in side the not so big train(t2) also travelling at the speed of light and so forth and still none of the trains would be going faster than c.

[glenc (Glen)]

Should I rename this thread Einstein's theory of relativity?

[leonie-macaroni (Leonie)]

Originally Posted by renormalised

The velocity of the particle (in this case bosons, or force carrying particles) does not dictate the mass of the particle.

Are you implying that if static, they remain massless? Or differ in mass?

Where can I learn more about particle physics?

I'm getting a bit lost, but this is all so interesting.