# B LIGO experiment Gravity waves

1. Mar 30, 2016

### carl susumu

I read that they found gravity waves that have the frequency of 35 to 250 hz but are not these frequencies that of sound waves? I thought sound does not propagate in vacuum or is there a new kind of gravity vacuum that contains a gravity ether or dark matter?

2. Mar 30, 2016

### Staff: Mentor

Sound waves can be in that frequency range, yes. But that does not mean any wave in that frequency range must be a sound wave.

No. Gravitational waves are waves of spacetime curvature.

3. Mar 30, 2016

### carl susumu

Do these gravity waves that have a frequency of 35 to 250 Hz make a sound, like a Chirp or something like that?

4. Mar 30, 2016

### Staff: Mentor

No. They're not sound waves.

5. Mar 30, 2016

### carl susumu

were Weber gravity wave sound waves?

6. Mar 30, 2016

### Staff: Mentor

No. The vibrations in Weber's bar detector that were triggered by gravitational waves, if indeed they were (most physicists believe they were actually not gravitational waves but measurement error), could have been considered "sound waves" (although they were much lower frequency than the frequency range you mentioned before). But those are not the same as the gravitational waves themselves (again, assuming his detector was actually detecting them, which it probably wasn't).

7. Mar 30, 2016

### secur

It's generally agreed that Weber didn't detect gravitational waves, he only thought he did. They wouldn't have been sound waves.

By the way the correct term is gravitational wave. A "gravity wave" is one which gets its energy from gravity. Best example, waves on the surface of the sea.

You probably saw some video where they played a "chirp" and said that was the sound made by LIGO's detected waves, right? No. The chirp was sound waves, generated by a conventional sound card, of the same frequency and duration as the gravitational waves.

8. Mar 30, 2016

### carl susumu

And what are the wavelengths of these so called gravity waves?

Last edited: Mar 30, 2016
9. Mar 30, 2016

### Staff: Mentor

Do you mean the gravitational waves that LIGO detected? The wavelengths of those waves are not given in the reports of the detection, probably because they are too long to be measured by the detector. The usual expectation for the wavelength of gravitational waves is that they will be comparable to the size of the source; the "size" in this case would be roughly the horizon radius of the black holes involved, which would be at least tens of kilometers.

No, gravitational waves are waves of spacetime curvature, as I said before.

Gravitational wave detectors like LIGO don't detect force, they detect strain, i.e., the fractional change in size of the detector. The strain of the waves that LIGO detected was around $10^{-21}$, i.e., the length of the arms of LIGO was changed by a fraction of roughly $10^{-21}$.

10. Mar 30, 2016

### carl susumu

A wavelength of five thousand miles! And what are the units of the strain, meters? Ha. Ha. Physicists are extremely funny bunnies.

Last edited: Mar 30, 2016
11. Mar 30, 2016

### secur

Strictly speaking they're so-called gravitational waves, but that's alright.

You mentioned the frequencies were 35 to 250 hz.
Speed of light is 3* 10^5 km / sec.
So wavelengths are 8600 to 1200 km.

But the sound wavelengths are much smaller at the same frequency.
The ratio: speed of sound / speed of light = 1.14 * 10^-6 (assuming dry air, 1 atmosphere pressure).
So sound wavelengths would be 7.54 to 2.2 meters.

The strain has no units, it's a dimensionless number, a ratio of lengths
- Hard to argue with that!

Last edited: Mar 30, 2016
12. Apr 1, 2016

### carl susumu

What are the lengths used to determine the strain?

13. Apr 1, 2016

### pervect

Staff Emeritus
The strain is a dimensionless quantity. Wiki has a a rather unclear definition, unfortunately, so I won't bother quoting it. If you consider that the base distance between the Ligo test masses is L, and that this distance changes by an amount $\Delta L$ due to the passing gravitational wave, the strain is $\frac{\Delta L}{L}$ which is a dimensionless quantity. For instance you might measure both L and $\Delta L$ in meters, or in centimeter, or in inches. Regardless of which units you use, the ratio, or strain, will be the same.

14. Apr 1, 2016

### carl susumu

Just one more question before I get back to the strain, is the length L decreasing or increasing?

15. Apr 1, 2016

### Staff: Mentor

It oscillates; the strain is the amplitude of the oscillation.