Exploring the Effects of Velocity on Energy

[Hernik]

Hi.

The higher the velocity of an object the more energy this object has. Velocity is purely relative. When I move I might *** well consider that I am still but all of my surroundings move. So if I travel with a velocity near the speed of light matter and radiation should appear more energetic in the direction of travel. Closing in on the moon it would seem so energetic, that it would maybe glow from heat. The visual light from stars in the direction I'm heading will appear and act on me as gamma radiation. This is right isn't it?

Now consider a photon on a path which should bring it close to a star on it's way to the earth. In the inertial frame of the photon the star closes in with the speed of light. As the star obviously has quite a lot of mass, the energy of the star must be infinite according to the photon. Energy curves space. Infinite energy must curve space infinitely? The photon should see the star as an enormous black hole. How can it ever get past the star - or any object with mass?

or maybe I should ask.. what's wrong with my reasoning?

Hope you can help me on this

 

[Dale]

The higher the velocity of an object the more energy this object has. Velocity is purely relative. When I move I might *** well consider that I am still but all of my surroundings move. So if I travel with a velocity near the speed of light matter and radiation should appear more energetic in the direction of travel. Closing in on the moon it would seem so energetic, that it would maybe glow from heat. The visual light from stars in the direction I'm heading will appear and act on me as gamma radiation. This is right isn't it?

Yes.

In the inertial frame of the photon the star closes in with the speed of light. ... what's wrong with my reasoning?

A photon does not have an inertial frame.

 

[bcrowell]

FAQ: What does the world look like in a frame of reference moving at the speed of light?

This question has a long and honorable history. As a young student, Einstein tried to imagine what an electromagnetic wave would look like from the point of view of a motorcyclist riding alongside it. But we now know, thanks to Einstein himself, that it really doesn't make sense to talk about such observers.

The most straightforward argument is based on the positivist idea that concepts only mean something if you can define how to measure them operationally. If we accept this philosophical stance (which is by no means compatible with every concept we ever discuss in physics), then we need to be able to physically realize this frame in terms of an observer and measuring devices. But we can't. It would take an infinite amount of energy to accelerate Einstein and his motorcycle to the speed of light.

Since arguments from positivism can often kill off perfectly interesting and reasonable concepts, we might ask whether there are other reasons not to allow such frames. There are. One of the most basic geometrical ideas is intersection. In relativity, we expect that even if different observers disagree about many things, they agree about intersections of world-lines. Either the particles collided or they didn't. The arrow either hit the bull's-eye or it didn't. So although general relativity is far more permissive than Newtonian mechanics about changes of coordinates, there is a restriction that they should be smooth, one-to-one functions. If there was something like a Lorentz transformation for v=c, it wouldn't be one-to-one, so it wouldn't be mathematically compatible with the structure of relativity. (An easy way to see that it can't be one-to-one is that the length contraction would reduce a finite distance to a point.)

What if a system of interacting, massless particles was conscious, and could make observations? The argument given in the preceding paragraph proves that this isn't possible, but let's be more explicit. There are two possibilities. The velocity V of the system's center of mass either moves at c, or it doesn't. If V=c, then all the particles are moving along parallel lines, and therefore they aren't interacting, can't perform computations, and can't be conscious. (This is also consistent with the fact that the proper time s of a particle moving at c is constant, ds=0.) If V is less than c, then the observer's frame of reference isn't moving at c. Either way, we don't get an observer moving at c.

 

[Hernik]

A photon does not have an inertial frame.

Thanks. Can you recommend somewhere where I can read more about this - easy on the math :-)... and thanks for the faq

 

[bcrowell]

Thanks. Can you recommend somewhere where I can read more about this - easy on the math :-)

For relativity without a lot of math, I like Relativity Simply Explained, by Martin Gardner.

 

[Hernik]

"Relativity simply explained". Got it, read it and loved it.

But it didn't explain why photons have no reference frames. Is that by convention?

- Henrik

 

[bcrowell]

"Relativity simply explained". Got it, read it and loved it.

Glad you liked it!

But it didn't explain why photons have no reference frames. Is that by convention?

See #3.

 

[Hernik]

Yes. I see...but I'm having a hard time trying to comprehend these concepts

Does post 3 boil down to:

- Because time and the distance traveled is always 0 for a photon it makes no sense to represent it in a coordinate system. And a reference frame refers to a coordinate system.

?

thanks, Henrik

 

[kj30]

Yes, I think the simplest "explanation" is that as you take relativistic effects to the limit of the speed of light, what the photon "observes" is that it takes zero time to reach its final destination, because it travels no distance (Lorentz contraction is total). As the photon "sees it", its destination is adjacent to its origin. It doesn't "experience" anything because it's a packet of energy frozen in its own time frame. It doesn't experience any time during which it would see stars going by.

 

[pervect]

A photon can have a coordinate system - or rather, one can create a coordinate system in which a photon has constant coordiantes. However, such a coordinate system is not an inertial coordinate system. It's sometimes useful, though abstract - such coordinates are known as "null coordinates", because the coordinates are neither timelike nor spacelike.

Suppose you have an ordinary 2d inertial frame (x,t). Then the coordinates
u = x+ct and v = x-ct define a coordinate system in terms of the null coordinates (u,v).

u will be constant for photons going in one direction, v will be constant for photons going in the other direction.

 

[Fredrik]

Does post 3 boil down to:

- Because time and the distance traveled is always 0 for a photon it makes no sense to represent it in a coordinate system. And a reference frame refers to a coordinate system.

No, because the phrase "for a photon" only makes sense if you have already found an appropriate coordinate system to think of as "the photon's reference frame". What it boils down to is that if you take a Lorentz transformation with speed v and let v→c, what you get isn't a Lorentz transformation, so this limit doesn't define an inertial coordinate system. It actually doesn't define any kind of coordinate system.

Yes, I think the simplest "explanation" is that as you take relativistic effects to the limit of the speed of light, what the photon "observes" is that it takes zero time to reach its final destination, because it travels no distance (Lorentz contraction is total). As the photon "sees it", its destination is adjacent to its origin. It doesn't "experience" anything because it's a packet of energy frozen in its own time frame. It doesn't experience any time during which it would see stars going by.

That explanation isn't correct. Massless particles don't have reference frames. This post explains it.

A photon can have a coordinate system - or rather, one can create a coordinate system in which a photon has constant coordiantes.

Yes, there are lots of coordinate systems in which a massless particle is stationary, and your example is the simplest. But I don't see a reason to call any of those coordinate systems "a photon's coordinate system".

 

[kj30]

That explanation isn't correct. Massless particles don't have reference frames. This post explains it.

That's why I had the word "explanation" in quotes. My point is you can already see from the basic SR equations taken to the limit v=c that the concept of "experiencing" something over time and distance as a photon doesn't make sense, just as the original poster observed that it doesn't make sense that other masses become infinite.

 

[lightarrow]

What if a system of interacting, massless particles was conscious, and could make observations? The argument given in the preceding paragraph proves that this isn't possible, but let's be more explicit. There are two possibilities. The velocity V of the system's center of mass either moves at c, or it doesn't. If V=c, then all the particles are moving along parallel lines, and therefore they aren't interacting, can't perform computations, and can't be conscious. (This is also consistent with the fact that the proper time s of a particle moving at c is constant, ds=0.) If V is less than c, then the observer's frame of reference isn't moving at c. Either way, we don't get an observer moving at c.

This is interesting, but how do you define the system's centre of mass for a massless particle?