Direction of time

[md65536]

9 hours ago, jajrussel said:

Ah, so the mirror Has to travel away from me at speed. Here I was jokingly thinking that my image photons should be sufficient😊. 
 

perhaps this means that if someone  has a mirror set up far enough away, I might see a reflection  half my age? Though maybe somewhat grainy?

The "perhaps" is right, the mirror doesn't have to be moving. If it's stationary relative to you it will only make you appear half your age at one specific age. If you set up a series of mirrors each 1/4 light years further away from you, then once every year, one of the mirrors would show you at half your current age. (1/4 LY away takes half a year for light to make a 2-way trip, so at age 1, light that left you half a year ago returns. At age 2 it takes 1 year return trip to a mirror 1/2 LY away, etc.)

7 hours ago, jajrussel said:

 it seems to suggest that if I am sitting in a train at near c looking at my reflection in a mirror held in my hand the top equation or the non-relativistic equation would suffice. However if the mirror is moving away from me at near c then the bottom equation becomes necessary. Could be that I am completely misunderstanding what it is saying, so I’ll switch to a bigger device (easier to read) and read it again.

If a mirror is moving away from you at any speed, your reflection is actually doubly Doppler shifted.

If you have any object moving away from you, and it is emitting light (eg. if it is self-lit or lit by a source in its own inertial frame) then it is red shifted according to the relativistic Doppler shift (even at non-relativistic speeds).

Symmetrically, an observer in the object's frame sees you red-shifted by the same amount. As for the reflection, it is equivalent to an observer receiving a red-shifted image of you, then projecting that image back to you, which you see as red-shifted again. So say it's a mirror with a wooden border and a light on it. The wood appears at an intensity and frequency that is sqrt((1-v/c)/(1+v/c)) times its emitted intensity and frequency. Your reflection appears at intensity and frequency of (1-v/c)/(1+v/c) times what a relatively stationary mirror shows.

The formulae will differ depending on whose frame you're talking about (and who has +v or -v in your setup), whether you're talking about source or observer, whether you're talking wavelength or frequency, etc (all these cases means is an inverse in the formula).

Edited October 9, 2019 by md65536

[md65536]

2 hours ago, md65536 said:

If it's stationary relative to you it will only make you appear half your age at one specific age.

Just to be clear, this is entirely due to delay time of light. Your reflection travels at a speed of c and doesn't age. Time dilation applies to a moving mirror's age, but that doesn't affect the timing (or speed) of the reflection.

 

As for the equations, sqrt((1-v/c)/(1+v/c)) = sqrt((c-v)/(c+v)) is the inverse Doppler factor, which is how many times the source frequency is multiplied to get what you see. (sqrt((c+v)/(c-v)) - 1) I think is the redshift, which is what fraction of the source wavelength is added to get what you see. Maybe someone can correct this or state it better. The source moving away from you has a positive velocity.

Edited October 9, 2019 by md65536

[jajrussel]

md65536 

😊———Thank you.

4 hours ago, md65536 said:

Just to be clear, this is entirely due to delay time of light. Your reflection travels at a speed of c and doesn't age. Time dilation applies to a moving mirror's age, but that doesn't affect the timing (or speed) of the reflection.

Ah... this would be the best example of time stopping at c. The reflection is what it is. I’m reasoning that my reflection can not age. (Actually, you said it) I’m just reasoning with it...

Your reflection travels at a speed of c and doesn't age. ( copied from your quote )

All it can do is show me a reasonable representation of the present/past dependent on conditions. In order for this to work. Time has to stop at c.

It might get a little weird if my reflection were to continue to age past my time.😊

Edited October 10, 2019 by jajrussel
Takes me a while to edit sometimes. If I cause confusion I’m sorry.

[md65536]

29 minutes ago, jajrussel said:

Ah... this would be the best example of time stopping at c. The reflection is what it is. I’m reasoning that my reflection can not age. All it can do is show me a reasonable representation of the present/past dependent on conditions. In order for this to work. Time has to stop at c.

It might get a little weird if my reflection were to continue to age past my time.😊

Sorry to have to confuse you, but the light that makes up an image, and the subject represented by the light, are two different things.

Suppose you print a photograph of yourself. The photo will age: discolor, disintegrate, etc. but that doesn't mean that it will show you aging in the image. In the case of beaming light across some great distance, neither would you age in the image (same as with a photo), nor would the light itself change with respect to time (as measured by an observer).

[jajrussel]

1 minute ago, md65536 said:

Sorry to have to confuse you, but the light that makes up an image, and the subject represented by the light, are two different things.

Suppose you print a photograph of yourself. The photo will age: discolor, disintegrate, etc. but that doesn't mean that it will show you aging in the image. In the case of beaming light across some great distance, neither would you age in the image (same as with a photo), nor would the light itself change with respect to time (as measured by an observer).

Not confused, sure the photograph will age, but the image is a capture of time stopped (the reflection). I’m just noting examples of time doing what it is supposed to do at c without the rest of the universe being effected. You see I used to be confused because it was difficult to think of time stopping as an effect of relativity while the person who’s time is seen to be stopped by an observer goes about his or her business completely unaware that someone has used relativity to decided that they the (observed) are frozen in time. Relativity is a lot easier for simple minds like mine to accept if the universe is allowed to continue on in spite of relativity’s predictions.

Actually, I am the one generally causing confusion. My mind moves around too much.

Relativity, doesn’t cause effects. What it does is explain the reasoning for what might be seen as illogical because what we see is just a reflection, or a type of radiation that can be affected by gravity or extreme speeds. Both cases need to be extreme, and even though we imagine the observed as not being effected in both cases they would be torn asunder( a condition necessary for the imagined activity.) luckily photographs and mirrors are more about the past and don’t require an unpleasant present.

[Mordred]

Try thinking of the scenario as signal delays the further the mirror is the longer it takes for your current image to arrive at the mirror. Now obviously if you place a mirror at each light year prior to your birth. Each year you age a year your birth image will arrive at each subsequent mirror. However it would the same amount of time for the reflection to arrive back to you.

 So for first mirror you wouldn't see the image until you were two. By then the image will arrive at the second mirror but would take two years to return so your would see the second image when you are 4. In the first mirror you would appear three. Etc etc.

A way to track this is take a graph paper.  At the top left corner set as the emitter. Each column represents a light year. Each row the following year record time time progression at each row column intersection.

A simple formula

[math]A_{seen}=A_{actual}- (2×d)[/math]

Edited October 10, 2019 by Mordred