#91

LeighH said:
Does it even make sense to ask that question though?

Let's mix that up a little. Image you jump into a spaceship in the Andromeda galaxy and set course for Earth at exactly the speed of light (somehow).

As you enter the Milky Way you pass a star and two of your clones climb into their identical spaceships and fly parallel to you.

You arrive on Earth and climb out to find two identical clones climbing out of theirs. This is the first time you have ever seen them. From your perspective, you have traveled instantaneously from Andromeda to Earth and would thus be completely unaware that your two clones accompanied you for some 50,000 years of your 2.5 million year journey.

Neither you nor your ship have actually aged a day.

Uh-hu! So red-shift doesn't occur at all?

#92

Capt' said:
The wavelength depends on the frame of reference , doppler shift and all that.

All three photons will be the same since the old and current one both have the same energy. E = h v

Unless you are talking about the view from the star in which case the question doesn't have any meaning since a view from another place would produce a different photon ( the earth rotates and orbits so being pedantic you would have to state the time and place you viewed the photon )

All three photons are the same? Seriously? In any frame of reference?

The three photons cannot be the same in any frame of reference.

#93

Capt' said:
The wavelength depends on the frame of reference , doppler shift and all that.

All three photons will be the same since the old and current one both have the same energy. E = h v

Oh! And by the way; red-shift does actually represent a loss of energy.

00:00
#94

himnextdoor said:
Oh! And by the way; red-shift does actually represent a loss of energy.
red-shift depends on your frame of reference.

If you match your velocity to that when the photon was emitted all three would be the same. If you don't match your velocity then sure you could accelerate in the opposite direction if you wanted to, or wait until the universe has expanded some more.

#95

Capt' said:
red-shift depends on your frame of reference.

If you match your velocity to that when the photon was emitted all three would be the same. If you don't match your velocity then sure you could accelerate in the opposite direction if you wanted to, or wait until the universe has expanded some more.

But all three photons travel at the speed of light regardless of the velocity of the emitter. There is one blue photon and two red photons; one of the red ones is an old blue one that has been red-shifted and has exactly the same wavelength of the younger red photon that has been travelling for a distance equal to the blue photon.

A detector would not be able to distinguish the age difference between the two red photons in any reference frame since the detector has the photons in its own frame. A comparison of three detectors might yield strange result but any one detector will see two of one and one of the other.

Why am I wrong?

Registered User
#96

It's all relative my friends.

It's a good point about not being able to tell the photons age.

Moderator
#97

himnextdoor said:
But it is a little redder.

Suppose a photon that has travelled 13-billion light-years passes by a star and finds itself on parallel paths with two other photons; the one on the left has a wavelength that matches the 'old' photon and the one on the right has a wavelength that matches the wavelength that the old photon had when it was first emitted.

If the old photon perceived no passage of time then wouldn't it have an identity crisis; which one would it be most like?

The photon would see itself as always having the same energy. Red-shifting photons are a feature of co-ordinates.

But all three photons travel at the speed of light regardless of the velocity of the emitter. There is one blue photon and two red photons; one of the red ones is an old blue one that has been red-shifted and has exactly the same wavelength of the younger red photon that has been travelling for a distance equal to the blue photon.

A detector would not be able to distinguish the age difference between the two red photons in any reference frame since the detector has the photons in its own frame. A comparison of three detectors might yield strange result but any one detector will see two of one and one of the other.

Why am I wrong?

What would the problem be? The detector will indeed see two of one and one of the other. The photons will all "agree" that the detector sees this, as the detector is not in the "reference frame" of any photon.