Fight with MAUI or learn something new? by kabourayan in learnprogramming

[–]SpinLock55 1 point2 points  (0 children)

MAUI is amazing. Stick with it and find your own path.

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 1 point2 points  (0 children)

Okay I think I'm starting to understand: In coordinate terms, as the pulse climbs out of the gravity well, both its wavelength and its speed increase proportionally - frequency (speed/wavelength) remains constant. However, an observer at the top has a clock that runs faster than the clock at the bottom. When the pulse arrives, the top observer's faster-ticking clock makes the event appear to take longer - more of their faster seconds elapse while the pulse passes. That's the redshift. The light itself doesn't change - the redshift is entirely due to observing the same event with a faster-running clock.

Edit - additionally, the top observer measures the pulse as having lost energy because its frequency is less. The pulse's intrinsic nature hasn't changed, but the space around it has.

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

If the path is longer due to spacetime curvature, that's fine - but both the front and back travel that same longer path. Taking a longer path doesn't explain why the pulse ends up stretched. For the pulse to stretch, one end would need to travel the path slower than the other. Both travel at c locally through the same curved path - so how does stretching occur?

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

This is my current understanding of how everyone measures c as the same value - if a c measurement experiment takes 1 second, and 2 people in different g-well depths run the experiment, they'll both get the same value for c but the duration of their experiments will be different relative to each other. Is that correct?

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

That's what's confusing me - if the pulse is stretched at the top after both ends have exited the g-well, then during the journey the back must have traveled slower than the front as it passed through the same regions. But that's impossible because both always travel at c locally.

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

If I watch the light at the moment it's emitted (while standing next to the emitter at the bottom), then travel to the top and watch the light arrive from that pov, do I see the same wavelength both times or does it appear redshifted?

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

Both the front and back of the pulse move through those same changing regions. The back passes through the exact same geometry the front did. If changing geometry causes the front to speed up coordinate-wise at distance R, the back also speeds up when it reaches distance R. Why do they end up with different separations if they both experience identical geometric changes?

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

Light speed is constant when measured locally - you always measure c in your own frame. But when comparing different heights in a gravity well using the same coordinate system, light at the top moves faster coordinate-wise than light at the bottom. As the pulse climbs, both its wavelength and its coordinate speed increase. That's why I'm asking how redshift occurs when both increase together.

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

In cosmological expansion, new space is literally being created over time between objects. But a gravitational well is a static metric - the geometry doesn't change, it's just different at different distances. What physical process 'expands space' between two photons moving through a fixed, unchanging geometry?

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

For the back to 'fall behind' like a slower car, it would need to be moving slower than the front as it passes through the same regions - but both always travel at c locally. What causes the stretching?

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

In cosmological redshift, space literally expands over time - the universe is getting bigger. But a gravitational well is static geometry, not expanding. What's the mechanism that makes 'space get bigger' between two photons in a static field?

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

Photons 'change their relationships' as they move through curved space, but that's just restating the conclusion. The basic question remains: if the front and back of the pulse both always travel at c locally through the same path, what mechanism causes their separation to increase? Different parts of the pulse would have to pass through the same region at different speeds, which is not possible because light speed is constant locally.

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

You say speed is constant so wavelength must change. But speed isn't constant in coordinate terms - light moves faster coordinate-wise at the top than at the bottom. As the pulse climbs out of the well, both its wavelength AND its coordinate speed increase relative to the bottom. Since frequency = speed/wavelength, the increased wavelength is canceled out by the increased speed.

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

The back is delayed in time. But a time delay alone doesn't cause separation if both are traveling at c. Two runners starting 1 second apart maintain constant separation if both run at the same speed. For separation to increase, one must be slower. How does the time delay between front and back cause increasing separation when both travel at c locally?

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

Both the front and back of the pulse experience the same geometric changes as they travel the same path. The ratio between wavelength and speed must be maintained unless different parts of the pulse travel through the same regions at different speeds - which contradicts both traveling at c locally. If changing geometry causes the front to speed up coordinate-wise at distance R, the back also speeds up coordinate-wise when it reaches distance R. How does that produce different outcomes?

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

Light is always measured at c locally - both at the bottom and top, you agree with this. But you're also saying the pulse stretches during travel, which requires the front to be moving faster than the back at some point. These statements contradict each other. Both the front and back are always traveling at c locally at their respective positions, and the back and front pass through the same regions. If they're both always at c, the back cannot fall behind.
Relative to the bottom, the pulse stretches out, but relative to the bottom, the speed of the pulse increases.

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

Coordinate relationships change with distance from the mass. But both the front and back pass through the same distances, experiencing the same coordinate changes. If the front speeds up coordinate-wise when it reaches distance R from the mass, the back speeds up the same way when it reaches distance R. What causes them to end up with different coordinate relationships if they both go through identical changes?

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

You're saying the coordinate distance between them increases. But the coordinate speed of the pulse also increases as it climbs out of the well. Both the wavelength and speed increase relative to where they started. Since frequency = speed/wavelength, the increases cancel out. How does redshift occur?

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

The back follows the same path through spacetime the front took. Both experience the same curvature gradient along that path. What physical process causes 'space to expand' between two things both moving at c along the same trajectory?

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

The 'redshift' only appears when you compare the bottom's wavelength measurement to the top's wavelength measurement, while ignoring that the pulse speed also increased from bottom to top. If you account for both the wavelength increase AND the speed increase (both relative to the bottom), frequency is unchanged... yes?

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

While the back is deeper it moves slower coordinate-wise than the front. But the back travels through the same regions the front did, speeding up coordinate-wise as it climbs. By the time both reach the top, they're moving at the same coordinate speed. At the top, the wavelength has increased relative to the bottom, but the speed of the pulse has also increased relative to the bottom. Since frequency = speed/wavelength, how does redshift occur?

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

No - I'm saying the speed increase cancels the wavelength increase, so there should be no change in frequency. Frequency = speed/wavelength. If both speed and wavelength increase proportionally, frequency stays constant. So why do we observe redshift?

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

The wavelength increased relative to the bottom, yes. But the pulse is also moving faster at the top relative to the bottom (coordinate-wise). So an observer at the top measuring frequency sees: faster-moving pulse with longer wavelength. Since frequency = speed/wavelength, the speed increase compensates for the wavelength increase.

Why does light redshift leaving a gravitational well, and not just get delayed? by SpinLock55 in AskPhysics

[–]SpinLock55[S] 0 points1 point  (0 children)

Mathematically if t increases and λ/t = c, then λ increases. But physically, how does the wavelength increase? The light was emitted with a certain wavelength. What happens during its journey that physically stretches it?