Two objects are identical, the second is hotter than the first. How to respond to the same strength?
using relativity
using relativity
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Let … L₀ = length of the two identical objects at the same low temperature T₁ …
… L = length of the two identical objects at the same high temperature T₂ …
Then … L = L₀ + α L₀ ΔT = L₀ [ 1 + α ΔT ] …… ΔT = T₂ ‒ T₁ …
… α = coefficient of linear expansion of the material …
Assume the object at lower temperature lying along the x-axis to be in the fixed
coordinate system (CS) …… the object at higher temperature also lying along the
horizontal to be in the CS moving with speed v relative to the fixed CS along the
x-axis. Then, the apparent length ……
…… L′ = L √ [ 1 ‒ v ² / c ² ] = L₀ [ 1 + α ΔT ] √ [ 1 ‒ v ² / c ² ]
of the object at the higher temperature as seen by an observer in the fixed CS is
shorter than its length L in its rest frame.
If the two objects now appear to have the same length in the fixed CS, then
…… L′ = L₀ ---> L₀ [ 1 + α ΔT ] √ [ 1 ‒ v ² / c ² ] = L₀ …… or ……
…… [ 1 + α ΔT ] √ [ 1 ‒ v ² / c ² ] = 1 ---> √ [ 1 ‒ v ² / c ² ] = 1 / [ 1 + α ΔT ] …
Solving for v ² / c ², we get ……
…… v ² / c ² = α ΔT [ 2 + α ΔT ] / [ 1 + α ΔT ] ²
……………. = 2 α ΔT [ 1 + ½ α ΔT ] [ 1 + α ΔT ] ⁻ ²
… L = length of the two identical objects at the same high temperature T₂ …
Then … L = L₀ + α L₀ ΔT = L₀ [ 1 + α ΔT ] …… ΔT = T₂ ‒ T₁ …
… α = coefficient of linear expansion of the material …
Assume the object at lower temperature lying along the x-axis to be in the fixed
coordinate system (CS) …… the object at higher temperature also lying along the
horizontal to be in the CS moving with speed v relative to the fixed CS along the
x-axis. Then, the apparent length ……
…… L′ = L √ [ 1 ‒ v ² / c ² ] = L₀ [ 1 + α ΔT ] √ [ 1 ‒ v ² / c ² ]
of the object at the higher temperature as seen by an observer in the fixed CS is
shorter than its length L in its rest frame.
If the two objects now appear to have the same length in the fixed CS, then
…… L′ = L₀ ---> L₀ [ 1 + α ΔT ] √ [ 1 ‒ v ² / c ² ] = L₀ …… or ……
…… [ 1 + α ΔT ] √ [ 1 ‒ v ² / c ² ] = 1 ---> √ [ 1 ‒ v ² / c ² ] = 1 / [ 1 + α ΔT ] …
Solving for v ² / c ², we get ……
…… v ² / c ² = α ΔT [ 2 + α ΔT ] / [ 1 + α ΔT ] ²
……………. = 2 α ΔT [ 1 + ½ α ΔT ] [ 1 + α ΔT ] ⁻ ²
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Sorry your question is not clear.
They are identical *when*? Before one is heated, or after?
[EDIT: Are they identical before, or after heating? In other words, are you asking if the extra energy of heat makes it harder to accelerate the heated mass? If so, yes it should.
We'd have to get it to 25 million degrees or so to have a hope of measuring the difference on our most sensitive scale, which would be hard on the object and the scale...]
They are identical *when*? Before one is heated, or after?
[EDIT: Are they identical before, or after heating? In other words, are you asking if the extra energy of heat makes it harder to accelerate the heated mass? If so, yes it should.
We'd have to get it to 25 million degrees or so to have a hope of measuring the difference on our most sensitive scale, which would be hard on the object and the scale...]