0
$\begingroup$

I was thinking about light and the different frequencies. The higher the frequency the more energetic the photon.

The higher frequency photon being more energetic seems counterintuitive when considering observed time dilation. What I mean is, if something has more energy it should "slow time" more than something with low energy. So (with my limited knowledge in the field) I would assume higher energy would be associated with a lower frequency because time is slower. Why is it that in nature this is the other way around?

$\endgroup$
1
$\begingroup$

It's probably best to not think of single photons as sources of gravitational energy. For one thing, most bulk electromagnetic fields are not eigenstates of the photon number operator.

For another thing, the thing that couples to the gravitational field is the energy density of the field. This density is proportional to the intensity of the field, rather than to its wavelength. It's true that it would take "more" red photons to make the same energy density than it would take blue photons, but once you've set up the energy density it doesn't matter, and a bulk electromagnetic field doesn't typically have a well-defined photon count, anyway.

This intuition, built up from classical field theory is actually part of why the photoelectric effect was so surprising when it was discovered, and why Einstein's explanation of it was so novel.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

I am not sure what you mean by

if something has more energy it should "slow time" more than something with low energy

But even if I just neglect that for a moment. I'll use a description of waves for visual aid and simplicity. Imagine in "not slowed down time" a wave has a period (and frequency) of 1 in this time. So there is one wave crest and one wave trough. If we then go to a slower time, by example a factor of 2 slower. It will take two times as long to reach the new unit of time.

So there would be two wave crests and troughs per unit of time. The period, in this new (and slower) unit of time, is now 0.5. Since frequency is $\frac{1}{period}$ the frequency will now be 2. So the frequency will be larger in the "slower time" than in the "normal time".

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.