Missing energy..

[blike]

More questions woohoo!

 

I always come up with a handful of questions while in class, but never can think of them when I'm at home. So I'll just post them as I think of them.

 

Frequency= © / (wavelength)

Energy=(planck's constant)(frequency)

 

So when the wavelength increases, the total energy decreases.

 

Now, here's my question..

 

Say a star is speeding away from the earth. If I'm traveling behind the star at the same speed as the star in my rocketship, my wavelength measurement will be lets say 600nm (just for example). Now, someone on earth who takes measurements gets the wavelength measurement of 800nm.

 

The earth's measurement of energy will be less than my rocketships. Where did the energy go? Which is the accurate measure of energy?

[Radical Edward]

not having done the maths, I'm not entirely certain. what I can say is that no energy should be lost, and I strongly suspect that the answer lies in momentum conservation, and of course the pervasive Lorentz Transforms. you might want to work through those and see if you can answer it for yourself, they aren't too complicated. I'll do it as well to see if I'm right or not ^^

[danny8522003]

If this thread is still answerable, i debated for a while with a friend of mine about this and came up with the following conclusion based upon his arguement.

 

Say the star is moving at x ms-1 will emit a wavelength per second as will a second object at 2x ms-1.

The first object will have a whole wave length emitted at these points.

. . . . . . . . . . . ->

The second at these points

. . . . . . . . . . . . . ->

 

Because the started at the same point the front end of the 2 photons will be in the same place. But because the final emitting point is in a different place it will mean the photon is more spread out and its wave length longer. The energy is all there. In this sense the space craft does not change the speed of the photon and nor does the star. But it does change the place from which it is emitted or received.

 

Concerning e=hf the frequency is lowered. Imagine how long it takes for one wavelength to pass you from end to end. The speed is constant. The longer wavelength will take longer to pass. By inversing this (1/time) you get the frequency. This leaves the energy lower. Basically e=hf will only measure the energy of a source assuming it is relatively still.

[swansont]

The answer is actually simpler. Energy is not invariant when you transform between frames. It's conserved only within a frame of reference.

[J.C.MacSwell]

The answer is actually simpler. Energy is not invariant when you transform between frames. It's conserved only within a frame of reference.

 

As an example the kinetic energy of an object disappears in it's own reference frame.