B Gravitational Redshift: Will Two Lasers Operate Differently?

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Two identical lasers placed at different altitudes will operate at different frequencies when measured from varying altitudes due to gravitational redshift and blueshift effects. The laser at sea level will appear to have a blueshift when measured from a higher altitude, while the mountain laser will exhibit a redshift when measured from sea level. When using identical local clocks, both lasers will read the same frequency. However, the mountain laser will be perceived as slower compared to clocks below it and faster compared to those above it. The discussion also touches on how redshift in cosmology is influenced by factors such as recession velocity and gravitational effects on light.
Torog
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If we run two identical lasers and put one at sea level and one on top of a high mountain, will they operate at different frequencies?
 
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They might depending on what altitude you're measuring their respective frequencies from. Measuring at an altitude different from the altitude of the respective laser will produce a red/blueshift. Measuring the frequency of the mountain laser from sea level will produce a blueshift, and measuring the frequency of the sea level laser from a higher altitude will produce a redshift.
 
Torog said:
If we run two identical lasers and put one at sea level and one on top of a high mountain, will they operate at different frequencies?
Adding on to what @Drakkith said, if you measure their frequencies with identical local clocks they will read the same.
 
What if I use the lasers as clocks - measure the pulses, divide many times and put the data to a readout. Shouldn't the clock (laser) on the mountain be slower?

I do understand that light as it falls into a gravitational field it gains energy and moves to blue and the opposite happens as light has to work its way out of a gravitational field.
 
Torog said:
Shouldn't the clock (laser) on the mountain be slower?
All of the above still applies. It will be slower than normal for any clock above it, faster for any clock below it, and unaltered for a clock right next to it.
 
Excuse me if I shift to cosmology. From what I understand the observed red shift of stars should come from three factors - according to the present model - First is a red shift given by the recession velocity (or general expansion of the Universe) second by the red shift caused by light having to make its way out of the gravitational field and third by the light having come from a star, quasar or other with a strong gravitational field where time and chemical processes run slower (redder) due to time being slower in the heavy gravitational field.
 
Torog said:
First is a red shift given by the recession velocity (or general expansion of the Universe) second by the red shift caused by light having to make its way out of the gravitational field and third by the light having come from a star, quasar or other with a strong gravitational field where time and chemical processes run slower (redder) due to time being slower in the heavy gravitational field.
The second and third are the same.
 
Torog said:
Excuse me if I shift to cosmology. From what I understand the observed red shift of stars should come from three factors - according to the present model - First is a red shift given by the recession velocity (or general expansion of the Universe) second by the red shift caused by light having to make its way out of the gravitational field and third by the light having come from a star, quasar or other with a strong gravitational field where time and chemical processes run slower (redder) due to time being slower in the heavy gravitational field.
We can combine redshift and gravitational time dilation like this:

$$ totalRedshift = \sqrt {redshift * gravitationalTimeDilation } $$

Or like this:

$$ totalRedshift = 0 * redshift + gravitationalTimeDilation $$

Or like this:

$$ totalRedshift = redshift + 0 * gravitationalTimeDilation $$That's a joke or something. But if we calculate the redshift factor and the gravitational time dilation factor, we always get the same number. So the formulas are correct, in a way.
 
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